If the length of a certain rectangle is decreased by 4 cm and the width is increased by 3 cm, a square with the same area as the original rectangle would result. The perimeter of the original rectangle (in cm) is :

Answer: Option D

Let the length and width of the rectangle be $$l$$ cm and b cm respectively.
Then,
$$\begin{array}{rl}& \left(l-4\right)\left(b+3\right)=lb\\ & \Rightarrow lb+3l-4b-12=lb\\ & \Rightarrow 3l-4b=\mathrm{12.....}(i)\\ & \text{And, }\\ & l-4=b+3\\ & \Rightarrow l-b=\mathrm{7.....}(ii)\end{array}$$
Multiplying (ii) by 4 and subtracting (i) from it, we get :
$$l$$ = 16
Putting $$l$$ = 16 in (ii), we get : b = 9
∴ Perimeter of the original rectangle :
$$\begin{array}{rl}& =2\left(l+b\right)\\ & =[2\left(16+9\right)]cm\\ & =50cm\end{array}$$